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I am confused about something. If (all what I will write are operators) $x$ is compatible with $p_y$ that means they have the same eigenvectors. However, $x$ is compatible with $y$ which means they have the same eigenvectors. That makes the eigenvectors of $p_y$ and $y$ are the same which makes them compatible. But it's not the case.

Where is the logic wrong?

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$x$ is compatible with $p_y$ that means they have the same eigenvectors.

That is perhaps not the best way to say it. The fact that $x$ and $p_y$ are compatible tells you that there exists at least one set of basis vectors which are simultaneously eigenvectors of both $x$ and $p_y$. Let's call this set $\{v\}$. Similarly, the fact that $x$ and $y$ are compatible means that there exists at least one set of basis vectors which are simultaneously eigenvectos of both $x$ and $y$. Let's call this set $\{w\}$. Nothing guarantees that $\{v\}$ and $\{w\}$ are the same set. Therefore, the statement that $y$ and $p_y$ are compatible is not necessarily true, and is as you know in fact not true.

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  • $\begingroup$ Thanks a lot. I think my confusion sprang from my misconception that if two observables are compatible, then all their eigenvectors are mutual. But now all makes sense again :) $\endgroup$ – stackarthur84 Dec 11 '14 at 6:05
  • $\begingroup$ Don't forget that if an answer actually answers your question it's standard procedure to mark it as accepted (click the check mark). This makes the answerer happy and let's others see that the question has been answered. $\endgroup$ – DanielSank Dec 11 '14 at 8:39
  • $\begingroup$ I didn't know that, but you got it ;) $\endgroup$ – stackarthur84 Dec 12 '14 at 7:42
  • $\begingroup$ @stackarthur84: Part of the way this site works is based on "reputation". More reputation allows you to do things like edit questions and answers. You get reputation for getting upvoted answers, accepted answers, etc. $\endgroup$ – DanielSank Dec 12 '14 at 10:36
  • $\begingroup$ @DanielSank Dear Daniel, it would be really awesome if you could help out with this post (no satisfactory answers by others). Best wishes, merry Christmas. $\endgroup$ – user929304 Dec 23 '14 at 12:50

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